Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is the functional inverse of $f(\theta) = \sin\theta\sqrt{\tan\theta}$? Or, equivalently, what is the inverse of $$f(\theta)=\sin^2\,\theta\tan\,\theta=\frac{\sin^3\,\theta}{\cos\,\theta}$$

It comes from a physics setup involving two equivalently massed and charged pith balls separated by a certain distance, and the equation simplifies to $q = 4L\sin\theta\sqrt{\pi\epsilon_0mg\tan{\theta}}$, where $\pi$, $g$, and $\epsilon_0$ are the obvious physical constants and $L$ and $m$ will be fixed. The question asked for $\theta$ in terms of $q$, however, so I'm wondering if there is a way to rearrange this. I can't seem to find anything on the internet, and Wolfram refuses to reveal the steps for its complex rearranged formula.

share|improve this question
    
(+1). Good question. –  The Chaz 2.0 Dec 22 '11 at 17:42
add comment

1 Answer

up vote 8 down vote accepted

I will assume you are interested in finding $\theta = f^{-1}(x)$ for $x \geq 0$ with the range $0 \leq \theta < \frac{\pi}{2}$. $$ x^2 = \left(f(\theta)\right)^2 = \sin^2(\theta) \tan(\theta) = \frac{\tan^3(\theta)}{1+\tan^2(\theta)} $$ Hence $\theta = \arctan(y(x))$, where $y$ is the positive root of $y^3 = x^2 (1 + y^2)$.

Using Cardano's formula: $$ \theta(x) = \arctan\left( \frac{1}{3} \left(x^2+\frac{\sqrt[3]{2} x^{10/3}}{\sqrt[3]{2 x^4+3 \left(\sqrt{12 x^4+81}+9\right)}}+\frac{\sqrt[3]{2 x^4+3 \left(\sqrt{12 x^4+81}+9\right)} x^{2/3}}{\sqrt[3]{2}}\right) \right) $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.